A230902 Positive numbers such that half of the set of divisors are of the form x^2 + x*y + y^2 (A003136) and half not (A034020).
2, 5, 6, 8, 11, 14, 15, 17, 18, 23, 24, 26, 29, 32, 33, 35, 38, 41, 42, 45, 47, 51, 53, 54, 56, 59, 62, 65, 69, 71, 72, 74, 77, 78, 83, 86, 87, 89, 95, 96, 98, 99, 101, 104, 105, 107, 113, 114, 119, 122, 123, 125, 126, 128, 131, 134, 135, 137, 141, 143, 146, 149, 152, 153, 155, 158, 159, 161, 162
Offset: 1
Keywords
Examples
Triangle read by rows in which row n lists the divisors of n begins: 1(0^2+0*1+1^2); 1(0^2+0*1+1^2), 2; 1(0^2+0*1+1^2), 3(1^1+1*1+1^2); 1(0^2+0*1+1^2), 2, 4(0^2+0*2+2^2); 1(0^2+0*1+1^2), 5; 1(0^2+0*1+1^2), 2, 3(1^1+1*1+1^2), 6; 1(0^2+0*1+1^2), 7(1^1+1*2+2^2); 1(0^2+0*1+1^2), 2, 4(0^2+0*2+2^2), 8; 1(0^2+0*1+1^2), 3(1^1+1*1+1^2), 9; 1(0^2+0*1+1^2), 2, 5, 10; 1(0^2+0*1+1^2), 11; 1(0^2+0*1+1^2), 2, 3(1^1+1*1+1^2), 4(0^2+0*2+2^2), 6, 12(2^2+2*2+2^2); 1(0^2+0*1+1^2), 13(1^2+1*3+3^2); 1(0^2+0*1+1^2), 2, 7(1^1+1*2+2^2), 14; 1(0^2+0*1+1^1), 3(1^11+1*1+1^2), 5, 15, i.e. a(1)=2, a(2)=5, a(3)=6, a(4)=8, a(5)=11, a(6)=14, a(7)=15.
Programs
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Maple
isA003136 := proc(n) local x,y ; for x from 0 do if x^2 > n then return false; end if; for y from 0 do if x^2+x*y+y^2 = n then return true; elif x^2+x*y+y^2 > n then break; end if; end do: end do: end proc: isA230902 := proc(n) local a36,a20,d ; a36 := 0 ; a20 := 0 ; for d in numtheory[divisors](n) do if isA003136(d) then a36 := a36+1 ; else a20 := a20+1 ; end if; end do: simplify( a36=a20) ; end proc: for n from 0 to 200 do if isA230902(n) then printf("%d,",n); end if; end do: # R. J. Mathar, Nov 08 2013
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Mathematica
A003136Q[n_] := Resolve[Exists[{x, y}, Reduce[n == x^2 + x*y + y^2, {x, y}, Integers]]]; okQ[n_] := With[{dd = Divisors[n]}, 2 Count[dd, _?A003136Q] == Length[dd]]; Select[Range[200], okQ] (* Jean-François Alcover, Jun 07 2024 *)
Extensions
Corrected by R. J. Mathar, Nov 08 2013
Comments